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Find the exact value of each of the remaining trigonometric functions of \( \theta \). Rationalize denominators when applicable, \( \sec \theta=-5 \), given that \( \sin \theta>0 \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. Bin \( \theta=\frac{2 \sqrt{6}}{5} \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. cos \( \theta=\square \) B. The function is undefined. (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B.

Ask by Howell Carrillo. in the United States
Feb 03,2025

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Answer

**A.** \( \cos \theta = -\frac{1}{5} \)

Solution

Given the information: - \( \sec \theta = -5 \) - \( \sin \theta > 0 \) Let's find the remaining trigonometric functions step by step. ### 1. Find \( \cos \theta \): Since \( \sec \theta \) is the reciprocal of \( \cos \theta \): \[ \cos \theta = \frac{1}{\sec \theta} = \frac{1}{-5} = -\frac{1}{5} \] **Answer:** **A.** \( \cos \theta = -\frac{1}{5} \) ### 2. Find \( \sin \theta \): Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \sin^2 \theta + \left(-\frac{1}{5}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{1}{25} = 1 \] \[ \sin^2 \theta = 1 - \frac{1}{25} = \frac{24}{25} \] \[ \sin \theta = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5} \] (Since \( \sin \theta > 0 \)) ### 3. Find \( \tan \theta \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} = -2\sqrt{6} \] ### 4. Find \( \csc \theta \): \[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{2\sqrt{6}}{5}} = \frac{5}{2\sqrt{6}} = \frac{5\sqrt{6}}{12} \] (Rationalized the denominator) ### 5. Find \( \cot \theta \): \[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-2\sqrt{6}} = -\frac{\sqrt{6}}{12} \] (Rationalized the denominator) ### Summary of All Trigonometric Functions: - \( \sin \theta = \frac{2\sqrt{6}}{5} \) - \( \cos \theta = -\frac{1}{5} \) - \( \tan \theta = -2\sqrt{6} \) - \( \csc \theta = \frac{5\sqrt{6}}{12} \) - \( \sec \theta = -5 \) - \( \cot \theta = -\frac{\sqrt{6}}{12} \) **Final Answer for \( \cos \theta \):** **A.** \( \cos \theta = -\frac{1}{5} \)

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To find the trigonometric functions given that \( \sec \theta = -5 \), we first note that \( \sec \theta \) is the reciprocal of \( \cos \theta \). Therefore, \( \cos \theta = -\frac{1}{5} \). Next, we can use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \sin \theta \): \[ \sin^2 \theta + \left(-\frac{1}{5}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{1}{25} = 1 \] \[ \sin^2 \theta = 1 - \frac{1}{25} = \frac{24}{25} \] \[ \sin \theta = \sqrt{\frac{24}{25}} = \frac{2 \sqrt{6}}{5} \] Since we are given that \( \sin \theta > 0 \), we take the positive root. Now, we can find the remaining trigonometric functions: - For \( \tan \theta \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2 \sqrt{6}}{5}}{-\frac{1}{5}} = -2 \sqrt{6} \] - For \( \csc \theta \) (the reciprocal of \( \sin \)): \[ \csc \theta = \frac{1}{\sin \theta} = \frac{5}{2 \sqrt{6}} = \frac{5 \sqrt{6}}{12} \] - For \( \cot \theta \) (the reciprocal of \( \tan \)): \[ \cot \theta = \frac{1}{\tan \theta} = -\frac{1}{2\sqrt{6}} = -\frac{\sqrt{6}}{12} \] Finally, here’s the complete list of exact values: - \( \sin \theta = \frac{2\sqrt{6}}{5} \) - \( \cos \theta = -\frac{1}{5} \) - \( \tan \theta = -2\sqrt{6} \) - \( \csc \theta = \frac{5\sqrt{6}}{12} \) - \( \sec \theta = -5 \) - \( \cot \theta = -\frac{\sqrt{6}}{12} \) So the answers for \( \cos \theta \) and the immediate requested functions are: A. cos \( \theta = -\frac{1}{5} \) Would you like to explore more about the interesting history of trigonometric functions or perhaps some problem-solving tips on how to best tackle similar problems in the future?

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